Question

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds?
AgBr: Ksp = 5.4 x 10-13

Ag2CO3: Ksp = 8.0 x 10-12

AgCl: Ksp = 1.8 x 10-10


Ag2CO3 < AgBr < AgCl
AgBr < Ag2CO3 < AgCl
AgBr < AgCl < Ag2CO3
AgCl < Ag2CO3 < AgBr

Answer 1

Answer:

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s²  (for compounds with 1:1  ratio).

It follows then  that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹²  with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of  the ratio of ions 2:1 in Ag2CO3,  so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3  = 2s x s = 2 s² =  8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl  ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰  ∴ s = √ 1.8 x 10⁻¹⁰   = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Answer 2

Answer:

5.4 x 10^-13

Explanation:

Edmentum says it is correct